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x-128=-128(cái này tự ấn máy tính nha viết hết ra lâu lắm)
x=-128+128
x=0
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x - 128 = \(\left(4\dfrac{20}{21}-5\right):\left(\dfrac{4141}{4242}-1\right):\left(\dfrac{636363}{646464}-1\right)\)
⇔ x - 128 = \(\left(\dfrac{104}{21}-5\right):\left(\dfrac{-1}{42}\right):\left(\dfrac{-1}{64}\right)\)
⇔ x - 128 = \(\left(\dfrac{-1}{21}\right):\left(\dfrac{-1}{42}\right):\left(\dfrac{-1}{64}\right)\)
⇔ x - 128 = -128
⇔ x = -128 + 128
⇔ x = 0
Vậy x = 0
\(x-128=\left(4\dfrac{20}{21}-5\right):\left(\dfrac{4141}{4242}-1\right):\left(\dfrac{636363}{646464}-1\right)\)
\(\Rightarrow x-128=\left(-\dfrac{1}{21}\right):\left(-\dfrac{1}{42}\right):\left(-\dfrac{1}{64}\right)\)
\(\Rightarrow x-128=2:\left(-\dfrac{1}{64}\right)\)
\(\Rightarrow x-128=-128\)
\(\Rightarrow x=\left(-128\right)+128\)
\(\Rightarrow x=0\)
<=> x - 128 = -25/21 : (-1/64)
<=> x - 128 = 1600/21
<=> x = 128 + 1600/21
=> x = 4288/21
Ta có : \(x-128=\left(4\frac{20}{21}-5\right)\left(\frac{4141}{4242}-1\right):\left(\frac{636363}{646464}-1\right)\\ =>x-128=\left(-\frac{1}{21}\right):\left(-\frac{1}{42}\right):\left(-\frac{1}{64}\right)\\ =>x-128=-128\\ =>x=0\)
a, Ta có x - 128 =( \(4\frac{20}{21}-5\)):\(\left(\frac{4141}{4242}-1\right):\left(\frac{636363}{646464}-1\right)\)
\(\Rightarrow\)x-128= \(\left(\frac{104}{21}-5\right):\left(\frac{41.101}{42.101}-1\right):\left(\frac{63.10101}{64.10101}-1\right)\)
\(\Rightarrow\)x-128=\(\left(\frac{-1}{21}\right):\left(\frac{-1}{42}\right):\left(\frac{-1}{64}\right)\)
\(\Rightarrow x-128=-128\)
\(\Rightarrow x=\left(-128\right)+128\)
\(\Rightarrow x=0\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
a, \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=-\frac{11}{4}\)
\(\frac{1}{2}-x=\frac{57}{28}\)
\(x=-\frac{43}{28}\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b, \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow\left(2x-1\right)^2=5^2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy ...
a) \(-\frac{5}{7}-\left(\frac{1}{2}-x\right)=\frac{-11}{4}\)
\(\Rightarrow\left(\frac{1}{2}-x\right)=\left(-\frac{5}{7}\right)+\frac{11}{4}\)
\(\Rightarrow\frac{1}{2}-x=\frac{57}{28}\)
\(\Rightarrow x=\frac{1}{2}-\frac{57}{28}\)
\(\Rightarrow x=-\frac{43}{28}\)
Vậy \(x=-\frac{43}{28}.\)
b) \(\left(2x-1\right)^2-5=20\)
\(\Rightarrow\left(2x-1\right)^2=20+5\)
\(\Rightarrow\left(2x-1\right)^2=25\)
\(\Rightarrow2x-1=\pm5\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=5+1=6\\2x=\left(-5\right)+1=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6:2\\x=\left(-4\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{3;-2\right\}.\)
d) \(\frac{x-6}{4}=\frac{4}{x-6}\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=4.4\)
\(\Rightarrow\left(x-6\right).\left(x-6\right)=16\)
\(\Rightarrow\left(x-6\right)^2=16\)
\(\Rightarrow x-6=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}x-6=4\\x-6=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+6\\x=\left(-4\right)+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{10;2\right\}.\)
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